Let $(\Omega, \mathcal{F}, \mathbf{P})$ be a probability space, $X \in \mathcal L^1(\textbf{P})$, $t \in \mathbb R$ and $Y= X \vee t$.
We want to prove that:
$ \textbf{E}(X \mid Y ) = Y \mathbb{1}_{Y>t} +\textbf{E}(X \mid X\leq t) \mathbb{1}_{Y=t} $ , for all $ \omega \in \Omega$.
My thoughts so far:
We can write the random variable $X$ as $X= X\mathbb{1}_{X>t} + X\mathbb{1}_{X \leq t}$.
We can notice that $Y=X$ on the set $\{X>t\} = \{X \vee t = X\}=\{Y>t \} $, so $X\mathbb{1}_{X>t}=Y\mathbb{1}_{Y>t}$, and also that $\{X \leq t\}= \{X \vee t = t \} = \{Y=t \}$ , so $\mathbb{1}_{X\leq t}=\mathbb{1}_{Y=t}$.
We have that:
$ \textbf{E}(X \mid Y)= \textbf{E}(X\mathbb{1}_{X>t} + X\mathbb{1}_{X \leq t} \mid Y) = \textbf{E}(X\mathbb{1}_{X>t} \mid Y) + \textbf{E} (X\mathbb{1}_{X \leq t} \mid Y) = \\ =\textbf{E}(Y\mathbb{1}_{Y>t} \mid Y) + \textbf{E} (X\mathbb{1}_{X \leq t} \mid Y)= Y\mathbb{1}_{Y>t} + \textbf{E} (X\mathbb{1}_{X \leq t} \mid Y)$
Now, we need to show that
$\textbf{E} (X\mathbb{1}_{X \leq t} \mid Y) = \textbf{E}(X \mid X\leq t) \mathbb{1}_{Y=t}$
which is equivalent with showing that
$\textbf{E} (X\mathbb{1}_{X \leq t} \mid Y) = \textbf{E}(X \mid X\leq t) \mathbb{1}_{X \leq t} = \textbf{E}(X \mathbb{1}_{X \leq t}\mid X\leq t)$ ,
but I had a problem to continue.
Can you please tell me if there is any mistake in the proof and give me a hint on how to finish it?